我正在使用Redux进行状态管理。 如何将存储重置为初始状态?

例如,假设我有两个用户帐户(u1和u2)。 想象下面的一系列事件:

用户u1登录到应用程序并做了一些事情,所以我们在存储中缓存一些数据。 用户u1退出。 用户u2无需刷新浏览器即可登录应用。

此时,缓存的数据将与u1关联,我想清理它。

当第一个用户注销时,如何将Redux存储重置为初始状态?


当前回答

简单回答一下丹·阿布拉莫夫的问题:

const rootReducer = combineReducers({
    auth: authReducer,
    ...formReducers,
    routing
});


export default (state, action) =>
  rootReducer(action.type === 'USER_LOGOUT' ? undefined : state, action);

其他回答

我已经创建了清除状态的操作。因此,当我分派登出动作创建者时,我也分派动作来清除状态。

用户记录动作

export const clearUserRecord = () => ({
  type: CLEAR_USER_RECORD
});

注销操作创建器

export const logoutUser = () => {
  return dispatch => {
    dispatch(requestLogout())
    dispatch(receiveLogout())
    localStorage.removeItem('auth_token')
    dispatch({ type: 'CLEAR_USER_RECORD' })
  }
};

减速机

const userRecords = (state = {isFetching: false,
  userRecord: [], message: ''}, action) => {
  switch (action.type) {
    case REQUEST_USER_RECORD:
    return { ...state,
      isFetching: true}
    case RECEIVE_USER_RECORD:
    return { ...state,
      isFetching: false,
      userRecord: action.user_record}
    case USER_RECORD_ERROR:
    return { ...state,
      isFetching: false,
      message: action.message}
    case CLEAR_USER_RECORD:
    return {...state,
      isFetching: false,
      message: '',
      userRecord: []}
    default:
      return state
  }
};

我不确定这是否是最佳的?

From a security perspective, the safest thing to do when logging a user out is to reset all persistent state (e.x. cookies, localStorage, IndexedDB, Web SQL, etc) and do a hard refresh of the page using window.location.reload(). It's possible a sloppy developer accidentally or intentionally stored some sensitive data on window, in the DOM, etc. Blowing away all persistent state and refreshing the browser is the only way to guarantee no information from the previous user is leaked to the next user.

(当然,作为共享计算机上的用户,你应该使用“私人浏览”模式,自己关闭浏览器窗口,使用“清除浏览数据”功能,等等,但作为开发人员,我们不能期望每个人都总是那么勤奋)

只需编辑声明约简的文件

import { combineReducers } from 'redux';

import gets from '../';

const rootReducer = (state, action) => {
  let asReset = action.type === 'RESET_STORE';

  const reducers = combineReducers({
    gets,
  });

  const transition = {
    true() {
      return reducers({}, action);
    },
    false() {
      return reducers(state, action);
    },
  };
  return transition[asReset] && transition[asReset]();
};

export default rootReducer;

我发现Dan Abramov的回答很适合我,但它触发了ESLint no-param-reassign错误- https://eslint.org/docs/rules/no-param-reassign

下面是我如何处理它,确保创建一个状态的副本(这是,在我的理解,Reduxy的事情要做…):

import { combineReducers } from "redux"
import { routerReducer } from "react-router-redux"
import ws from "reducers/ws"
import session from "reducers/session"
import app from "reducers/app"

const appReducer = combineReducers({
    "routing": routerReducer,
    ws,
    session,
    app
})

export default (state, action) => {
    const stateCopy = action.type === "LOGOUT" ? undefined : { ...state }
    return appReducer(stateCopy, action)
}

但是也许创建一个状态的副本,然后把它传递给另一个减速器函数,它会创建一个状态的副本,这有点过于复杂了?这篇文章读起来不太好,但更切题:

export default (state, action) => {
    return appReducer(action.type === "LOGOUT" ? undefined : state, action)
}

这种方法非常正确:销毁任何特定状态“NAME”以忽略并保留其他状态。

const rootReducer = (state, action) => {
    if (action.type === 'USER_LOGOUT') {
        state.NAME = undefined
    }
    return appReducer(state, action)
}